We add excess NaCl solution (58.44 g/mol) to

56 mL of a solution of silver nitrate (AgNO3
169.88 g/mol), to form insoluble solid AgCl.
When it has been dried and weighed, the
mass of AgCl (143.32 g/mol) is found to be
1.25 grams.
What is the molarity of the original AgNO3
solution? The formula weight of NaNO3 is
85.00 g/mol.

See previous problems.

To find the molarity of the original AgNO3 solution, we can use stoichiometry and the concept of moles.

First, let's calculate the number of moles of AgCl formed by using the mass of AgCl that was obtained (1.25 grams) and the molar mass of AgCl (143.32 g/mol). To do this, we use the formula:

Moles = Mass / Molar mass

Moles of AgCl = 1.25 g / 143.32 g/mol = 0.0087 moles

Next, we need to determine the stoichiometric relationship between AgNO3 and AgCl. From the balanced chemical equation:

AgNO3 + NaCl -> AgCl + NaNO3

We can see that 1 mole of AgNO3 produces 1 mole of AgCl.

Since the moles of AgNO3 and AgCl are equal, the moles of AgNO3 in the original solution will also be 0.0087 moles.

Now, let's calculate the volume of the original AgNO3 solution (in liters) using the formula:

Molarity = Moles / Volume

Moles of AgNO3 = 0.0087 moles
Molarity of AgNO3 = Unknown
Volume of AgNO3 = Unknown

To determine the volume, we rearrange the formula:

Volume = Moles / Molarity

Substituting the known values:

Volume = 0.0087 moles / Molarity

Now, we need to convert the volume from milliliters (mL) to liters (L). We are given that the volume of the original AgNO3 solution is 56 mL:

Volume = 56 mL / 1000 mL/L = 0.056 L

Substituting this into the equation:

0.056 L = 0.0087 moles / Molarity

Now, solve for Molarity:

Molarity = 0.0087 moles / 0.056 L ≈ 0.155 M

Therefore, the molarity of the original AgNO3 solution is approximately 0.155 M.